Solve the linear system:
#"Equation 1":# #y=3x-2#
#"Equation 2":# #14x-3y=0#
The solution is the point #(x,y)# that the two lines have in common, which is the point of intersection. I'm going to use substitution to solve the system.
Equation 1 is already solved for #y#. Substitute #3x-2# for #y# in Equation 2 and solve for #x#.
#14x-3(3x-2)=0#
Expand.
#14x-9x+6=0#
Simplify.
#5x+6=0#
Subtract #6# from both sides.
#5x=-6#
Divide both sides by #5#.
#x=-6/5# or #-1.2#
Substitute #-6/5# for #x# in Equation 1. Solve for #y#.
#y=3(-6/5)-2#
Expand.
#y=-18/5-2#
Multiply #2# by #5/5# to get an equivalent fraction with #5# as the denominator.
#y=-18/5-2xx5/5#
#y=-18/5-10/5#
Simplify.
#y=-28/5# or #-5.6#
The solution is #(-6/5,-28/5)# or #(-1.2,-5.6)#.
graph{(y-3x+2)(14x-3y+0)=0 [-6.366, 4.73, -8.243, -2.696]}
